# Convert a cell with structures into a matrix

1 view (last 30 days)
Alejandro Fernández on 30 Jan 2021
Edited: Stephen on 30 Jan 2021
Hi, does anyone know how to could I go from the information I have stored in the variable A to what I have in the variable matrix?
% Input data.
A = cell(3,1);
A{1}.data = [1 2 3];
A{1}.a = 0;
A{2}.data = [4 5 6];
A{2}.a = 1;
A{3}.data = [7 8 9];
A{3}.a = 3;
% Result to obtain.
matrix = [1 2 3; 4 5 6; 7 8 9];

David Hill on 30 Jan 2021
m=[];
for k=1:3
m=[m;A{k}.data];
end
Stephen on 30 Jan 2021

Stephen on 30 Jan 2021
Edited: Stephen on 30 Jan 2021
Rather than inefficiently storing lots of scalar structures in a cell array, you should just use one efficient non-scalar array, then your task is trivial:
S(1).data = [1 2 3];
S(1).a = 0;
S(2).data = [4 5 6];
S(2).a = 1;
S(3).data = [7 8 9];
S(3).a = 3;
M = vertcat(S.data) % this is all you need!
M = 3×3
1 2 3 4 5 6 7 8 9
Note that you can also convert that unfortunate cell array of scalar structures to one structure array:
S = [A{:}];

R2020b

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